Question

The maximum value of $12\sin \theta -9{\sin }^2\theta$ is-

• A. $3$
• B. $4$
• C. $5$
• D. $Noneofthese$

Answer 1

Answer: (A)

$3$

Let $f\left(\theta \right)$ be the given function,

$f\left(\theta \right)=12\sin \theta -9{\sin }^2\theta$

Differentiate the given function with respect to $\theta$,

$f^{\prime }\left(\theta \right)=12\cos \theta -18\sin \theta \cos \theta$

$=6\cos \theta \left(2-3\sin \theta \right)$

Put $f^{\prime }\left(\theta \right)=0$

$6\cos \theta \left(2-3\sin \theta \right)=0$

$6\cos \theta =0,2-3\sin \theta =0$

$\cos \theta =0,\sin \theta =\frac{2}{3}$

$\theta =\frac{\pi }{2},\theta ={\sin }^{-1}\frac{2}{3}$

Substitute the value of $\theta$ in the given function,

$f\left(\frac{\pi }{2}\right)=12\sin \frac{\pi }{2}-9{\sin }^2\left(\frac{\pi }{2}\right)$

$=12-9$

$=3$

$f\left({\sin }^{-1}\frac{2}{3}\right)=12\sin \left({\sin }^{-1}\frac{2}{3}\right)-9{\left(\sin \left({\sin }^{-1}\frac{2}{3}\right)\right)}^2$

$=12\left(\frac{2}{3}\right)-9{\left(\frac{2}{3}\right)}^2$

$=-4$

Therefore, the maximum value of the given function is 3.